Example 7-35. Evaluate the integral
∫∫∫
V
F(x, y, z)dV, where
F =xˆe 1 +xy ˆe 2 +eˆ 3
and dV =dxdydz is a volume element. The limits of integration are determined from
the volume bounded by the surfaces y=x^2 , y = 4, z = 0,and z= 4.
Solution From figure 7-25 the limits of integration can be determined by sketching an
element of volume dV =dx dy, dz and then summing these elements in the x−direction
from x= 0 to x=√y to form a parallelepiped. Next sum the parallelepiped in the
z−direction from z= 0 to z= 4 to form a slab. Finally, the slab can be summed in
y−direction from y= 0 to y= 4 to fill up the volume.
Figure 7-25.
Volume bounded by y=x^2 , and the planes y= 4, z = 0 and z= 4
One then has
∫∫∫
V
F·dV =
∫y=4
y=0
∫z=4
z=0
∫ x=√y
x=0
(xˆe 1 +xy ˆe 2 +ˆe 3 )dx dz dy
=
∫ 4
0
∫ 4
0
∫√y
0
[xˆe 1 +xy ˆe 2 +ˆe 3 ]dxdzdy
Perform the integrations over each vector component and show that
∫∫
V
F ·dV = 16 ˆe 1 +^128
3 ˆe^2 +
64
3 eˆ^3
